Precipitation and Sea Level Pressure (SLP)

Te(d^2)y A(l^2)en

2. precip and SLP means and variances: space, time, spacetime

The GRAND (space+time) mean of precipitation and SLP are : 4.74mm; 1010.97hPa.
mean(x)

The GRAND (space+time) variance of precipitation and SLP are: 12.63 (mm)^2; 2.37(hPa)^2
variance = (sum(x^2) - sum(x)^2)/N

The GRAND standard deviations are: 3.55mm; 1.54hPa
sqrt(var_n(x))

The SPATIAL variances of my TIME MEAN longitude section are: 5.76 (mm)^2; 1.16 (hPa)^2
var_n( total(x,2)/NMONS )

The TEMPORAL variances of my LONGITUDE MEAN time series are: 0.94 (mm)^2; 0.55 (hPa)^2
var_n( total(x,1)/NLONS )

3. the assignment section

1. Confirm that the time mean of the anomalies as defined above is 0
The above statement is confirmed by averaging the anomalies for each longitude grid.
2. Is the spatial mean of the anomalies (as defined above) 0?
The RED time-series plots below show that the spatial mean of the anomalies (longitude mean) is NOT zero.
Is it the same as the time series of the spatial mean of the raw data? Or is it a new thing?
The spatial mean of the anomalies is distinct from the actual spatial mean of the raw data.

TO COMPUTE THE BELOW TO PLOT:
spatial (longitude) mean of the entire data series lonmean_x = (total(x,1)/nlons)
spatial (longitude) mean of the anomaly lonmean_anomx = (total(anomx,1)/nlons)
difference between the series and anomaly diffx = (lonmean_x - lonmean_anomx)